A neutral pion, \(\pi^{0}\text{,}\) decays into two light particles, called photons. The rest mass of neutral pion is \(135\:\textrm{MeV/c}^2\text{.}\) That is, if you multiply \(135\:\textrm{MeV/c}^2\) by \(c^2\) you will get the rest energy in MeV unit. Find the energy and momenta of the two photons released when a neutral pion at rest decays into two photons.
Hint.
Use formulas in this section.
Answer.
\(1.13\times10^{-11}\: \textrm{J}\text{,}\) \(3.77\times 10^{-20}\:\textrm{kg.m/s}\text{.}\)
Solution.
Let \(E\) be the energy of one of the photons. Balancing the energy in the decay reaction we find
\begin{equation*}
2E = 135\:\textrm{MeV}.
\end{equation*}
Therefore, the energy of each photon is
\begin{align*}
E \amp = 67.5 \:\textrm{MeV} = 67.5 \:\textrm{MeV} \times 10^6\times 1.67\times 10^{-19}\: \textrm{J/MeV}\\
\amp = 1.13\times10^{-11}\: \textrm{J}.
\end{align*}
The momentum of each photon
\begin{equation*}
p = \frac{E}{c} = \frac{1.13\times10^{-11}\: \textrm{J}}{3.0\times 10^8\:\textrm{m/s}} = 3.77\times 10^{-20}\:\textrm{kg.m/s}.
\end{equation*}
